Question

How do you simplify $\dfrac{{{{\left( {2{x^3}{z^2}} \right)}^3}}}{{{x^3}{y^4}{z^2}.{x^{ - 4}}{z^3}}}$ and write it only using positive exponents?

Answer 1

Hint: We will first open the cube in numerator and remove the parentheses. Then, we will use the fact that ${u^a}{u^b} = {u^{a + b}}$ on both x and y and then use the fact that $\dfrac{{{x^a}}}{{{x^b}}} = {x^{a - b}}$.

Complete step-by-step solution:
We are given that we are required to simplify $\dfrac{{{{\left( {2{x^3}{z^2}} \right)}^3}}}{{{x^3}{y^4}{z^2}.{x^{ - 4}}{z^3}}}$ and write it only using positive exponents.
Now, we can write the given expression $\dfrac{{{{\left( {2{x^3}{z^2}} \right)}^3}}}{{{x^3}{y^4}{z^2}.{x^{ - 4}}{z^3}}}$ as written follows:-
$ \Rightarrow \dfrac{{{{\left( {2{x^3}{z^2}} \right)}^3}}}{{{x^3}{y^4}{z^2}.{x^{ - 4}}{z^3}}} = \dfrac{{{2^3}{{\left( {{x^3}} \right)}^3}{{\left( {{z^2}} \right)}^3}}}{{{x^3}{y^4}{z^2}.{x^{ - 4}}{z^3}}}$
Simplifying the right hand side of the above equation, we will then obtain the following equation with us:-
$ \Rightarrow \dfrac{{{{\left( {2{x^3}{z^2}} \right)}^3}}}{{{x^3}{y^4}{z^2}.{x^{ - 4}}{z^3}}} = \dfrac{{8{x^9}{z^6}}}{{{x^3}{y^4}{z^2}.{x^{ - 4}}{z^3}}}$
Now, we know that we have a fact given by the following expression with us:-
$ \Rightarrow {u^a}{u^b} = {u^{a + b}}$, where u, a and b are real numbers.
Therefore, we will now obtain the following equation with us:-
$ \Rightarrow \dfrac{{{{\left( {2{x^3}{z^2}} \right)}^3}}}{{{x^3}{y^4}{z^2}.{x^{ - 4}}{z^3}}} = \dfrac{{8{x^9}{z^6}}}{{{x^{3 - 4}}{y^4}{z^{2 + 3}}}}$
Simplifying the powers on the right hand side, we will then obtain the following equation with us:-
$ \Rightarrow \dfrac{{{{\left( {2{x^3}{z^2}} \right)}^3}}}{{{x^3}{y^4}{z^2}.{x^{ - 4}}{z^3}}} = \dfrac{{8{x^9}{z^6}}}{{{x^{ - 1}}{y^4}{z^5}}}$
Now, we know that, we have a fact given by the expression: $\dfrac{{{x^a}}}{{{x^b}}} = {x^{a - b}}$
Therefore, we have the following equation with us now:-
$ \Rightarrow \dfrac{{{{\left( {2{x^3}{z^2}} \right)}^3}}}{{{x^3}{y^4}{z^2}.{x^{ - 4}}{z^3}}} = \dfrac{{8{x^{9 - ( - 1)}}{z^{6 - 5}}}}{{{y^4}}}$
Simplifying it further, we will get:-
$ \Rightarrow \dfrac{{{{\left( {2{x^3}{z^2}} \right)}^3}}}{{{x^3}{y^4}{z^2}.{x^{ - 4}}{z^3}}} = \dfrac{{8{x^{10}}z}}{{{y^4}}}$
Thus, we have an expression with only positive exponents.

Note: The students must note that 0 is not a positive number. It is just a non – zero number but not a positive one. Positive number is strictly greater than zero. Therefore, any positive number can never be equal to zero. Therefore, we just remove the x portion with the 0 exponent because that would not have been classified as a positive exponent.
The students must also keep in mind that while using the identity that says that $\dfrac{{{x^a}}}{{{x^b}}} = {x^{a - b}}$, one must make sure of the fact that $x \ne 0$ and $x \in \mathbb{R}$. Here, in the given question, we knew that x is not equal to 0, because if x would have been equal to 0, then the expression itself would have become zero in the initial only.
The students must also note that we have used the fact that ${\left( {{x^a}} \right)^b} = {x^{ab}}$.